Problem: $\dfrac{dy}{dx}=\dfrac{3x^2}{y^2}$ Which curve solves the differential equation and passes through the point $(0,2)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ y=\sqrt[3]{x^3+8}$ (Choice B) B $ y=\sqrt[3]{3x^3+2}$ (Choice C) C $ y=\sqrt[3]{x^3+2}$ (Choice D) D $ y=-\sqrt[3]{3x^3+2}$ (Choice E) E $ y=\sqrt[3]{3x^3+8}$
Solution: The differential equation is separable. $\dfrac{dy}{dx}=3x^2\cdot\dfrac1{y^2}$ What does it look like after we separate the variables? $y^2\,dy = 3x^2\,dx$ Let's integrate both sides of the equation. $\int y^2\,dy = \int 3x^2\,dx$ What do we get? $\dfrac13y^3=x^3 + C$ What value of $C$ makes the solution curve pass through the point $(0,2)$ ? Let's substitute $x=0$ and $y=2$ into the equation and solve for $C$. $\begin{aligned} \dfrac13\cdot2^3 &= 0^3 + C\\ \\ \\ \dfrac13\cdot8 &=0+C\\ \\ \\ C&=\dfrac83 \end{aligned}$ Now use this value of $C$ to express $y$ in terms of $x$. $\begin{aligned} \dfrac13y^3&=x^3 + \dfrac83\\ \\ \\ y^3&=3x^3 + 8\\ \\ y&=\sqrt[3]{3x^3+8} \end{aligned}$